Suppose we have cube of 1.00 cm length. It is cut in all three directions, so as to produce eight cubes, each 0.50 cm on edge length. Then suppose these 0.50 cm cubes are each subdivided into eight cubes 0.25 cm on edge length, and so on. How many of these successive subdivisions are required before the cubes are reduced in size to colloidal dimensions of 100 nm.

Suppose we have cube of 1.00 cm length. It is cut in all three directi

1.	Suppose we have cube of 1.00 cm length. It is cut in all three directions, so as to produce eight cubes, each 0.50 cm on edge length. Then suppose these 0.50 cm cubes are each subdivided into eight cubes 0.25 cm on edge length, and so on. How many of these successive subdivisions are required before the cubes are reduced in size to colloidal dimensions of 100 nm.
Answer» Solution :We find that every division in two equal halves also reduces the size of edge LENGTHS to one half. In first subdivision 1 cm is reduceds to `0.5 cm = (1)/(2) cm.` In second subdivision 0.5 cm is reduced to `0.25 cm = (1)/(4) cm = ((1)/(2))^(2) cm` In n subdivision 1 cm is reduced to `((1)/(2))^(n)`. Size of colloidal particles lies between 1 to 1000 mm. Thus to make n subdivision required PARTICLE size may be ATTAINED. `((1)/(2))^(n) = 100 nm = 100 xx 10^(-9) m = 100 xx 10^(-7) cm.` n log 2 = 5 `n xx 0.3010 = 5.` `n = (5)/(0.3010) = 16.61 = 17` subdivisions are required for dimension of 100 nm.